Somebody help me with algreba

Find the quotient of (11 ^ 18)/(11 ^ 3)

astraxan [27]

Answer:

The first choice, $11^{15}$ .

Step-by-step explanation:

The numerator, $11^{18}$ , is equivalent to eighteen " $11$ " multiplied together:

$11^{18} = \underbrace{11 \times 11 \times 11 \times 11 \times \cdots \times 11 \times 11}_{\text{$18$ in total}}$ .

On the other hand, the denominator, $11^{3}$ , is equivalent to three " $11$ " multiplied with one another:

$11^{3} = 11 \times 11 \times 11$ .

Dividing $11^{18}$ by $11^{3}$ would eliminate three " $11$ " from the numerator:

$\begin{aligned}& \frac{11^{18}}{11^{3}} \\ =\; & \frac{(11 \times 11 \times 11) \times \overbrace{11 \times \cdots \times 11}^{\text{$(18 - 3)$ in total}}}{(11 \times 11 \times 11)} \\ =\; & \overbrace{11 \times \cdots \times 11}^{\text{$15$ in total}} \\ =\; & 11^{15}\end{aligned}$ .

In general, dividing an expression by $y^{b}$ ( $y \ne 0$ ) is equivalent to multiplying that expression by $y^{-b}$ .

For example, in this question, dividing $11^{18}$ by $11^{3}$ would be equivalent to multiplying $11^{18}\!$ by $11^{-3}$ . In other words:

$\begin{aligned}& \frac{11^{18}}{11^{3}} \\ =\; & 11^{18} \times 11^{-3} \\ =\; & 11^{18 - 3} \\ =\; & 11^{15}\end{aligned}$ .

6 0

3 years ago

1.Parallelogram ACFD is split in two parts so that ABED and FEBC are congruent isosceles trapezoids. What are the measures of al

DiKsa [7]

Refer to the diagram shown below.

Because ACFD is a parallelogram, its opposite angles are equal. Therefore
x = m∠ACF = m∠BCF = 48°
Similarly,
y = m∠CAD = m∠CFD

The sum of the angles inside a parallelogram is 360°. Therefore
48° + x + y + y = 360°
Because x = 48°,
48° + 48° + 2y = 360°
2y = 360° - 96° = 264°
y = 132°

Because ABED and FEBC are congruent, therefore
y = m∠DAB = m∠CFE = 132°
x = m∠ADE = m∠FCB = 48°

Because FEBC is a parallelogram, the opposite angles are equal. Therefore
m∠CBE = m∠CFE = y = 132°
m∠BCF = m∠BEF = x = 48°

Answer:
The measures of all angles of trapezoid FEBC are
m∠BCF = 48°
m∠BEF = 48°
m∠CBE = 132°
m∠CFE = 132°